How do you find the integral of #(e^x)(cosx) dx#?

Question

Emma Aldridge · Accepted Answer

I found: #(e^x(cos(x)+sin(x)))/2+c#

I tried Integration by Parts (twice) and a little trick...!

Evelyn Agard · Answer

#int\ e^xcos(x)\ dx=e^x/2(sin(x)+cos(x))+C#

Alternatively, we can use a nice little technique called complexifying the integral.

We notice that #cos(x)# is just the same as the real part of #e^(ix)# (by Euler's identity, #e^(itheta)=cos(theta)+isin(theta)#). We can use this fact to rewrite the integral like so: #int\ e^xcos(x)\ dx=int\ e^x*Re(e^(ix))\ dx=#

In terms of complex numbers, #e^x# is just some real factor, so it doesn't matter whether we have it outside or inside the Real part function. This means we can put the entire integral inside the Real part function: #=Re(int\ e^xe^(ix)\ dx)=Re(int\ e^(x+ix)\ dx)=Re(int\ e^((i+1)x))\ dx=#

We can do a quite simple u-substitution to evaluate the integral: #Re(e^((i+1)x)/(i+1)+C)=Re((e^(x)e^(ix))/(i+1))+C=#

#=e^x*Re(e^(ix)/(i+1))+C=#

We can now use Euler's identity again to expand the top. We also multiply by the conjugate of the bottom to simplify the fraction: #=e^x*Re((i-1)/((i+1)(i-1))(cos(x)+isin(x)))+C=#

#=e^x*Re((i-1)/(-1-1)(cos(x)+isin(x)))+C=#

#=e^x*Re((i/(-2)-1/(-2))(cos(x)+isin(x)))+C=#

#=e^x*Re(-i/2cos(x)+1/2sin(x)+1/2cos(x)+i/2sin(x))+C=#

We can now quite easily pick out the real parts: #=e^x(1/2sin(x)+1/2cos(x))+C=e^x/2(sin(x)+cos(x))+C#

Emily Ammerman · Answer

undefined To find the integral of $ e^x \cdot \cos(x) \, dx $, you can use integration by parts. Let $ u = e^x $ and $ dv = \cos(x) \, dx $. Then, $ du = e^x \, dx $ and $ v = \sin(x) $. Using the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substituting the values:

$$ \int e^x \cdot \cos(x) \, dx = e^x \sin(x) - \int \sin(x) \cdot e^x \, dx $$

This results in another integral similar to the original one. You can solve it by using integration by parts again, or you can recognize that the integral of $ e^x \sin(x) $ can be found by using integration by parts in a similar manner as before. After solving, you'll find the antiderivative of $ e^x \cos(x) $.